JPS6251073B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a power supply circuit that connects a power supply transformer and a load transformer with a line to supply power.

In conventional power supply circuits, especially distribution systems, when the capacity of distribution transformers increases as the load increases, the leakage impedance of the transformer decreases and short-circuit current increases. There was a drawback that it had to be replaced with one with larger breaking capacity. A known method for suppressing short-circuit current is to increase the leakage impedance of a transformer. However, increasing the leakage impedance of a transformer increases the voltage drop of the transformer due to load current, and also reduces the voltage drop due to leakage flux. The transformer case may locally overheat due to eddy currents generated in the transformer case, etc., or mechanical thrust may act on the transformer windings in the event of a short circuit failure, damaging the windings, or causing damage to the transformer windings.
There was a drawback that the capacity of the next winding became large.

SUMMARY OF THE INVENTION An object of the present invention is to provide a power supply circuit that can reduce leakage impedance of a transformer and suppress short-circuit current to a small value in order to reduce voltage drop in the transformer due to load current. It is in.

The power supply circuit of the present invention will be explained in detail below with reference to the drawings.

First, in order to explain the principle of the present invention, consider the single-phase power supply circuit shown in FIG. In the same figure, T
_s is a power distribution transformer using a three-legged iron core l having a central leg l ₀ and both side legs l ₁ and l ₂ , and a side leg l ₁ of the iron core l has a first primary winding 1 s and A first secondary winding 2s is wound around the side leg _l2 , and a second primary winding 1's and a second secondary winding 2's are wound around the side leg l2. In the figure, the first primary winding and the secondary winding are shown as being wound on different parts of the side leg l1 of the iron core, but in reality they are wound on different parts of the side leg _l1 of the iron core. These windings are wound concentrically on the same part of the side leg l ₁ . The same applies to the second primary winding and secondary winding.

The first and second primary windings of the transformer T _S are connected in series, and a power supply G is connected to both ends of the series circuit of these primary windings. Note that Z _T indicates the leakage impedance of the transformer T _S. The impedance of the power supply G is omitted for simplicity. One ends of the first and second secondary windings 2s and 2's of the power supply side transformer T _S are connected to each other via an impedance Z, and the other ends of both secondary windings are commonly connected. There is. Also, both ends of the first secondary winding 2s are connected to a line 3,
3' to a load Z _A , and both ends of the second secondary winding 2's are connected to a load Z _B via lines 4, 4'.

In the circuit shown in Figure 1, it is assumed that the first and second primary windings 1s and 1's and the first and second secondary windings 2s and 2's of the transformer T _S have the same number of turns. . Therefore, when there is no load, a voltage of 1/2 of the power supply voltage E is induced in all four windings of the transformer T _S .

Next, let the currents of the loads Z _A and Z _B be I _A and I _B , respectively, and the current flowing through the impedance Z be I _Z. Also, the currents flowing in the secondary windings 2s and 2's are respectively I ₂ and I' ₂ , and the currents flowing in the primary windings 1s and 1's are
Let I be the voltage flowing through. When the current of each part is determined in this way, the following equation is established according to Kirchhoff's current law.

Next, since the excitation current of the transformer T _S is small compared to the load current, if this is ignored, the sum of the magnetomotive forces of the windings 1s and 2s will be zero according to the principle of a transformer.
Similarly, the sum of the magnetomotive forces of windings 1's and 2's also becomes zero. From this, the following equation is established.

I=I ₂ =I' ₂ (2) The following equation is obtained from equations (1) and (2).

Calculating I and I _Z from equation (3), I = (I _A + I _B )/2 ......(4) I _Z = (I _B - I _A )/2 ......(5) Next, 2 Let the terminal voltages of the next windings 2s and 2's be V _A and V _B. Since the number of turns of the primary winding 1s and the secondary winding 2s are equal, according to the principle of a transformer, the primary winding 1s
The terminal voltage of s becomes V _A , and similarly the primary winding 1's
The terminal voltage of is _VB . Therefore, by applying Kirchhoff's voltage law to the primary circuit, the following equation can be obtained.

E=IZ _T +V _A +V _B ......(6) Next, the secondary windings 2s and 2's are connected to the impedance Z
Applying Kirchhoff's voltage law to the closed circuit formed by connecting in parallel via , the following equation is obtained.

V _A −I _Z Z−V _B =0 (7) When V _A and V _B are calculated from equations (4) to (7), the following equations are obtained.

V _A = (E/2) - (1/4) (I _A + I _B ) Z _T - (1/4) (I _A - I _B ) Z ...... (8) V _B = (E/2) −(1/4)(I _A +I _B )Z _T +(1/4)(I _A −I _B ))Z……(9) As is clear from equations (8) and (9), I When _A = I _B , the third term on the right side becomes zero, and V _A and V _B become irrelevant to impedance Z. Therefore, the voltage drop due to the load current is due to the second term on the right side of equations (8) and (9), and is only due to the leakage impedance Z _T of the transformer.

As mentioned above, the leakage impedance Z _T of the transformer T _S decreases as the windings are wound so that the leakage impedance between the primary and secondary windings becomes small, and as the transformer capacity increases, The voltage drop due to load current becomes smaller as the transformer capacity increases. Note that the impedance Z is the load current I _A and I _B
Since it acts on the difference in current, a small capacity one is sufficient.

Next, consider a case where a short circuit failure occurs in the line 3 in the circuit shown in FIG. In order to avoid mixing the load current and short-circuit current, it is assumed that the failure occurs when there is no load, and Z _A = 0 and Z _B = ∞. Since Z _A =0, V _A =0, and since Z _B =∞, I _B =
It becomes 0. Substituting this relationship into equation (8) yields the following equation.

(E/2) - (1/4) I _A Z _T - (1/4) I _A Z = 0
......(10) When the short circuit current I _A is determined from equation (10), it becomes I _A = (2E) / (Z _T + Z) ...... (11). As is clear from equation (11), by increasing the impedance Z, the short-circuit current I _A can be suppressed to a small value.

Next, a method for selecting impedance Z will be explained. If the percent impedance of the transformer T _S is %Z _T , its ohmic impedance Z _T is given by the following equation.

Z _T (10% Z _T V ² )/P [Ω] ………(12) However, here, P is the rated capacity of the transformer [KVA]
, and V indicates the rated voltage [KV] of the transformer. Percent impedance of transformer %Z _T
is generally constant regardless of the transformer capacity, so
The ohmic impedance Z _T of the transformer is obtained from equation (12),
It is inversely proportional to the transformer capacity P. Therefore, as the transformer capacity P increases, the ohmic impedance Z _T decreases.

Let us now consider the case where the distribution transformer T _S with a rated capacity P is replaced with a rated capacity mP which is m times P (m>1). From equation (12), the ohmic impedance Z′ _T of a transformer with rated capacity mP is given by the following equation.

Z′ _T = (10%Z _T V ² )/(mP) = (1/m) Z _T
......(13) Here, Z _T is the ohmic impedance of the transformer with rated capacity P. That is, when the capacity of a transformer increases by m times, its ohmic impedance decreases to 1/m. The short-circuit current of the conventional power supply circuit is I _A = (2E) / Z _T (14) when Z in equation (11) is set to zero, and in the circuit of the present invention, the capacity of the transformer is
The short circuit current when doubled is I _A = (2E)/{(Z _T /m) + Z} (15) from equation (11). Therefore, when finding the value of impedance Z such that both are equal from equations (13) and (14), it becomes: Z={1-(1/m)}Z _T (16). In other words, if the impedance Z is selected to the value of equation (16), the magnitude of the short circuit current before and after replacing the transformer will be the same, so the circuit breaker that was used before replacing the transformer will be the same after replacing the transformer. can be used as is.

Next, consider the magnitudes of voltages V _A and V _B when impedance Z is selected to the value of equation (16). Substituting Z′ _T in equation (13) and Z in equation (16) for Z _T and Z in equations (8) and (9), equations (8) and (9) become (17), respectively.
and (18).

V _A = (E/2) - (1/4) (I _A + I _B ) Z _T + (1/2m) (m-1) I _B Z _T ...... (17) V _B = (E/2 ) − (1/4) (I _A + I _B ) Z _T + (1/2 m) (m-1) I _A Z _T ………(18) Since m>1 here, (17), The third terms on the right side of equation (18) are both positive, which cancels out the voltage drop in the second term. The first term on the right side of equations (17) and (18) is the voltage at no load, and the second term is the voltage drop of the transformer before replacing the transformer, so adding the third term to this is not possible. , which means that the voltage drop after replacing the transformer will be smaller than that before replacing the transformer. As described above, according to the present invention, the voltage drop due to load current can be reduced.

Next, the extreme unbalanced state of the two loads Z _A and Z _B ,
That is, with no load on one side, Z _B =∞, so I _B =0.
Consider the case of In this case, equations (17) and (18) become equations (19) and (20), respectively.

V _A = (E/2) - (1/4) I _A Z _T ...... (19) V _B = (E/2) - (1/4) I _A Z _T + (1/2 m) (m -1) I _A Z _T ...... (20) As can be seen from equations (19) and (20), do not make V _B higher than the voltage E/2 when both Z _A and Z _B are unloaded. In order to do so, it is necessary to satisfy the following formula.

−(1/4)I _A Z _T +(1/2m)(m-1)I _A Z _T ≦
0......(21) Therefore, m≦2......(21)' In other words, when replacing a distribution transformer with one with twice the rated capacity or less, one of the two loads Even in an extreme unbalanced state where ZA and ZB are unloaded, the voltage on the no-load side can be suppressed to below the voltage E/2 when both Z _A and Z _B are unloaded.
Further, by configuring the load curves of the two loads to match as much as possible, it is possible to suppress the voltage increase on the light load side even when m exceeds 2.

Although FIG. 1 shows a single-phase power supply circuit, FIG. 2 shows the configuration when the present invention is applied to a three-phase power supply circuit. In Fig. 2, ts ₁ , ts ₂ , and ts ₃ are three unit transformers configured in the same way as the transformer shown in Fig. 1. By combining these three unit transformers, the power supply side transformer can be created. Vessel T
_S is configured. First and second primary windings 1s connected in series of each of three unit transformers
and 1's are triangularly connected to form the 3-phase primary winding on the power supply side.
Ws ₁ is configured. 3 unit transformers TS ₁ ~
The first secondary winding 2s of ts ₃ is three-phase triangularly connected to form the first power supply side three-phase secondary winding _Ws2 , and similarly the second secondary winding 2's is triangularly connected. A second power supply side three-phase secondary winding W's ₂ is configured. The output terminals of the first and second power supply rule three-phase secondary windings Ws ₂ and W′s ₂ are respectively connected to the first and second three-phase bus B ₁
and _B2 , and the lines L _A to L are connected to the first bus _B1 .
_C via the first load Z _A and also the second bus B ₂
A second load _ZB is connected to each of the lines L' _A to L' _C via lines L'A to L'C. Further, terminals of corresponding phases of the first power supply side three-phase secondary winding Ws ₂ and the second power supply side three-phase secondary winding W's ₂ are connected to each other via an impedance Z. If the configuration shown in FIG. 2 is rewritten to make it easier to understand, it will become as shown in FIG. 3.

Next, consider a single-phase power supply circuit shown in FIG. 4 as a modification of the present invention. In this circuit, the common connection point of the secondary windings 2s and 2's in FIG. 1 is directly grounded. In the circuit shown in Fig. 4, the ground fault current I _A when a ground fault occurs in the line 3 is given by the equation (11) above.
Similarly to the formula, the following formula is obtained.

I _A = (2E) / (Z _T +Z) ...... (22) As can be seen from equation (22), it can be seen that by increasing the impedance Z, the ground fault current can be suppressed to a small value. Next, the ground voltage V _B of the sound line in this case is determined by the equation (9) with I _B = 0, that is, V _B = (E/2) - (1/4) I _A Z _T + (1 /4) I _A Z
......(23) becomes. Therefore, if the impedance Z is selected as Z=Z _T (24), V _B becomes as follows.

V _B =E/2 (25) Therefore, V _B becomes the voltage E/2 during no-load, and the voltage of the sound line of the line 4 will not rise when the line 3 has a ground fault. Here, in equation (22), Z
Since I _A when = 0 is the ground fault current in the conventional direct grounding power transmission system, it can be seen that when Z = Z _T the ground fault current is 1/2 of that in the conventional system. In short, if the impedance Z is selected to be equal to the leakage impedance of the transformer (consider including the power source impedance when there is an impedance of the power source _G ) Z 1/2, and in this case the voltage to ground of the healthy line does not rise.

Although FIG. 4 shows a single-phase power supply circuit, when this configuration is applied to a three-phase parallel two-line power supply circuit, it becomes as shown in FIG. 5. The circuit of FIG. 5 corresponds to the power source transformer T _S in which the secondary side of FIG. 3 is star-connected and the neutral point is grounded. The load-side transformer T _R has the same configuration as the power-side transformer T _S , only the primary and secondary transformers are switched. In this case, the first and second power supply side primary windings W _R1 and W' _R1 of the load side transformer T _R are star-connected, and their respective neutral points are directly grounded. With this configuration, the load current flows through the two parallel lines L _A to L _C and L' _A to L' _C , and no current flows through the impedance Z. Therefore, the voltage drop due to the load current becomes independent of the impedance Z. Further, the ground fault current and short circuit current are suppressed to small values by the impedance Z.

Next, in the embodiment shown in FIG. 2, if the impedance Z is configured by a series circuit of a capacitor C and a reactor L, the voltage on the line on the overload side can be increased and the voltage on the line on the light load side can be decreased. Can be done. However, since this increases the short circuit current, it is necessary to short circuit the capacitor C in the event of a short circuit failure in order to prevent this.

Further, as shown in FIG. 6, a gap 5 is provided in the center leg of the three-leg iron core of the transformer, and a winding 6 is wound around this center leg, and this winding 6 has an impedance Z.
It can also be used as This method is economical since it is not necessary to separately manufacture the impedance Z.

Furthermore, as shown in Fig. 7, the first two parts of the transformer
It is also possible to connect an impedance Z between the intermediate tap P of the secondary winding 2s and the intermediate tap P' of the second secondary winding 2's. This is economical because the insulation of the impedance Z can be reduced.

As described above, according to the present invention, when increasing the capacity of a transformer connected to a power distribution system as the load increases, it is possible to reduce the voltage drop due to the load current, and also to reduce the short-circuit current. Since the current value can be suppressed to a certain value, there is no need to replace the circuit breaker used before replacing the transformer with one with a larger breaking capacity, and there is an advantage that the circuit breaker can be used as is. Furthermore, by selecting the impedance Z to an appropriate value and keeping the increase in the capacity of the distribution transformer to less than double, there is an advantage that the voltage increase on the lighter load side of the two loads can be suppressed. Furthermore, if the load curves of the two loads are similar, the voltage increase on the light load side can be suppressed even if the capacity of the transformer increases by more than twice. In addition, by increasing the capacity of the transformer and reducing leakage magnetic flux, it is possible to not only suppress heat generation due to eddy currents generated in the transformer case, etc., but also damage the windings due to axial thrust acting on the transformer windings in the event of a short-circuit failure. It is possible to prevent this from happening.
Furthermore, since equal load current flows through the two secondary windings of the transformer, the capacities of the primary and secondary windings can be made equal, increasing the capacity of the secondary windings. It is economical because it is not necessary.

[Brief explanation of the drawing]

Fig. 1 is a block diagram showing an embodiment in which the present invention is applied to a single-phase power supply circuit, Fig. 2 is a block diagram showing an embodiment in which the present invention is applied to a three-phase power supply circuit, and Fig. 3 is a block diagram showing an embodiment in which the invention is applied to a three-phase power supply circuit. A circuit diagram showing only the electrical configuration of the embodiment shown in the figure, FIG. 4 is a block diagram showing another embodiment in which the present invention is applied to a single-phase power supply circuit, and FIG. 5 is an embodiment of the embodiment shown in FIG. A circuit diagram showing the electrical configuration of an example system in which a load-side transformer is added to a three-phase parallel two-line power transmission system, and FIG. 6 is a configuration showing a modified example of the transformer used in the present invention. 7 are configuration diagrams showing essential parts of another embodiment of the present invention. T _S ...power supply side transformer, l ... three-leg iron core, l ₀ ...
...Central leg, l ₁ , l ₂ ...Side leg, 1s...First primary winding, 1's...Second primary winding, 2s...First
Secondary winding, 2's...Second secondary winding, Z...
Impedance, G... power supply, Z _A , Z _B ... load, T _R ... load side transformer, 1 _R ... first primary winding, 1' _R ... second primary winding, 2 _R ...first 2
Secondary winding, 2' _R ...Second secondary winding.

Claims (1)

[Claims] 1. In a power supply circuit that connects a power source to a load via a transformer, the transformer includes a three-leg iron core and a first primary wound around one side leg of the three-leg iron core. a second primary winding and a second secondary winding wound around the other side leg of the three-legged iron core; the primary windings are connected in series and connected to a power source; the first and second secondary windings of the transformer are connected in parallel via an impedance; the first and second secondary windings of the transformer A power supply circuit characterized in that the windings are each connected to a load. 2. The impedance is connected between one end or intermediate point of the first secondary winding and one end or intermediate point of the second secondary winding, and the other ends of both secondary windings are grounded. The power supply circuit according to claim 1, characterized in that: 3. In a power supply circuit that connects a power source to a load via a transformer, the transformer includes a three-leg iron core and a first primary winding and a secondary winding wound around one side leg of the three-leg iron core. three unit transformers each having a second primary winding and a secondary winding wound around the other side leg of the three-leg iron core; The first and second primary windings of the three unit transformers are connected in series, and the series-connected first and second primary windings of the three unit transformers are connected in three-phase connection. The first secondary winding and the second secondary winding of the three unit transformers are connected to a three-phase power supply, respectively, to form the first and second three-phase secondary windings. the terminals or midpoints of corresponding phases of the first and second three-phase secondary windings are respectively connected via impedance, and the first and second three-phase secondary windings A power supply circuit, each of which is connected to a load. 4. The power supply circuit according to claim 3, wherein the first and second three-phase secondary windings are triangularly connected. 5. The power supply circuit according to claim 3, wherein the first and second three-phase secondary windings are star-connected and each neutral point is grounded.